Question

How do you write the linear function f with the values $f(2) = - 1$and $f(5) = 4$?

Answer 1

Hint: In order to determine the linear function, suppose that the $({x_1},{y_1}) = (2, - 1)$and $({x_2},{y_2}) = (5,4)$, putting these values in the slope point form of a line $m = \dfrac{{y{ _2} - {y_1}}}{{{x_2} - {x_1}}}$ to obtain the slope of the linear function. Put the value of slope in the slope intercept form of a line $y = mx + C$ and determine the value of C by substituting the 2 for x and -1 for y. Putting back the C and slope in the slope intercept form you will get your required function.

Complete step-by-step solution:
We are given a linear function $f$ with the values $f(2) = - 1$ and $f(5) = 4$. From the given we can find that when the value of variable in the function is equal to 2 it gives the answer equal to -1 and similar when variable equal to 5 answers is 4.
Let the variable be $x$and $f(x) = y$
Assume that $({x_1},{y_1}) = (2, - 1)$ and $({x_2},{y_2}) = (5,4)$
Since the function is a linear function, so the slope point form of the line
$\Rightarrow m = \dfrac{{y{ _2} - {y_1}}}{{{x_2} - {x_1}}}$
Putting the values, we get
 $
\Rightarrow m = \dfrac{{4 - ( - 1)}}{{5 - 2}} \\
 \Rightarrow m = \dfrac{{4 + 1}}{3} \\
 \Rightarrow m = \dfrac{5}{3} \\
 $
The slope of the function comes to be $m = \dfrac{5}{3}$.
Let’s substitute the value of slope in the slope-intercept form of the line
$\Rightarrow y = mx + C$
$\Rightarrow y = \dfrac{5}{3}x + C$---------(1)
Now to determine the value of C, substitute the 2 for x and -1 for y and find out the value of C
$
\Rightarrow - 1 = \dfrac{5}{3}\left( 2 \right) + C \\
\Rightarrow C = - \dfrac{{13}}{3} \\
 $
Substituting the value of C in the equation (1),we get
$\Rightarrow y = \dfrac{5}{3}x - \dfrac{{13}}{3}$
$\Rightarrow f\left( x \right) = y = \dfrac{5}{3}x - \dfrac{{13}}{3}$

Therefore, the required linear function is $f\left( x \right) = y = \dfrac{5}{3}x - \dfrac{{13}}{3}$.

Additional Information:
i) Linear Equation: A linear equation is a equation which can be represented in the form of $ax + c$where $x$is the unknown variable and a,c are the numbers known where $a \ne 0$.If $a = 0$then the equation will become constant value and will no more be a linear equation.
ii) The degree of the variable in the linear equation is of the order 1.
iii) Every Linear equation has 1 root.

Note:
1. The slope (also called as gradient) of every straight line shows the sleepiness and the direction of the line.
2.Remember the graph of every linear function is always a straight line.
3. Slope-intercept form is only for linear functions.
4.You can cross check your result by putting the values of $x = 2\,and\,x = 5$in the equation obtained, the result will be $ - 1\,and\,4$respectively.
When $x = 2$
$
 \Rightarrow y = \dfrac{5}{3}\left( 2 \right) - \dfrac{{13}}{3} \\
 \Rightarrow y = \dfrac{{10 - 13}}{3} = - 1 \\
 $
And when $x = 5$
$
 \Rightarrow y = \dfrac{5}{3}\left( 5 \right) - \dfrac{{13}}{3} \\
 \Rightarrow y = \dfrac{{25 - 13}}{3} = \dfrac{{12}}{3} \\
  \Rightarrow y = 4 \\
 $
Hence, the result obtained was correct.